Tag: hyperbola

Quizzes (10) Questions (269)

Questions Related to hyperbola

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The second-degree curve and pair of asymptotes differ by a constant. Let the second-degree curve $S = 0$ represent the hyperbola then respective pair of asymptote is given by.$\displaystyle S+\lambda =0\left ( \lambda \in R \right )$ which represent a pair of straight lines so $\lambda$  can be determined. The equation of asymptotes is $\displaystyle A=s+\lambda =0$ if equation of conjugate hyperbola of the curve $S =0$ be represents by $S _{1}$, then $A$ is arithmetic mean of the curves $S _{1}$, & $ S $.

A hyperbola passing through origin has $\displaystyle 2x-y+3=0$ and $\displaystyle x-2y+2=0$ as its asymptotes, then equation of its transverse and conjugate axes are:

  1. $\displaystyle x-y+2=0$ and $\displaystyle 3x-3y+5=0$
  2. $\displaystyle x+y+2=0$ and $\displaystyle 3x-3y+5=0$
  3. $\displaystyle x-y+1=0$ and $\displaystyle 3x-3y+5=0$
  4. $\displaystyle2 x-2y+1=0$ and $\displaystyle 3x-3y+5=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The transverse axis of hyperbola is the bisector of the angle between the asymptotes containing the origin and the conjugate axis is the other bisector.


 And equation of bisector of angle of the asymptotes are given by

$\displaystyle \frac{2x-y+3}{\sqrt{5}}=\pm \frac{x-2y+2}{\sqrt{5}}$

$\displaystyle \Rightarrow  2x-y+3 =\pm \left ( x-2y+2 \right )$

$\displaystyle \Rightarrow  2x-y+3 =x-2y+2$

and $\displaystyle  2x-y+3 =x-2y+2= -\left ( x-2y+2 \right )$

$\displaystyle  \Rightarrow x+y+1=0 \ and \ 3x-3y+5=0$

Hence, option 'C' is correct.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of a hyperbola have equations $y-1=\dfrac{3}{4}(x+3).$ If a focus of the hyperbola has coordinates $(7,1)$, the equation of the hyperbola is

  1. $\dfrac{(x+3)^2}{16}-\dfrac{(y-1)^2}{9} = 1$
  2. $\dfrac{(y-1)^2}{9}-\dfrac{(x+3)^2}{16} = 1$
  3. $\dfrac{(x+3)^2}{64}-\dfrac{(y-1)^2}{36} = 1$
  4. $\dfrac{(y-1)^2}{36}-\dfrac{(x+3)^2}{64} = 1$
  5. $\dfrac{(x+3)^2}{4}-\dfrac{(y-1)^2}{3} = 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation of asymptotes are 

$y-1=\dfrac { 3 }{ 4 } (x+3  )$    ......(i)

$ y-1=-\dfrac { 3 }{ 4 } (x+3)$     .....(ii)

Centre of the hyperbola is point of intersection of asymptotes.

Therefore, by solving (i) and (ii), we get centre as $C(-3,1)$.

Slope of asymptotes $=\dfrac { b }{ a } $

$\Rightarrow \dfrac { b }{ a } =\pm \dfrac { 3 }{ 4 }$      ......(i)

Focus is $(7,1)$.

Focus for hyperbola of form $\dfrac { { (x-h) }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { (y-k) }^{ 2 } }{ { b }^{ 2 } } =1$ is $(h+ae,k)$

$\Rightarrow 7=-3+ae\\ \Rightarrow ae=10\\ \Rightarrow a\dfrac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } }  }{ a } =10\\ \Rightarrow \sqrt { { a }^{ 2 }+{ b }^{ 2 } } =10$

Substituting $b$ from (i), we get

$\Rightarrow \sqrt { { a }^{ 2 }+{ \left( \pm a \dfrac { 3 }{ 4 }  \right)  }^{ 2 } } =10\\ \Rightarrow \dfrac { 5a }{ 4 } =10\\ \Rightarrow a=8\\ \Rightarrow b=\pm \dfrac { 3 }{ 4 } a=\pm 6$

So, the equation of hyperbola is

$\dfrac { { (x+3) }^{ 2 } }{ { 8 }^{ 2 } } -\dfrac { { (y-1) }^{ 2 } }{ { 6 }^{ 2 } } =1$

$\dfrac { { (x+3) }^{ 2 } }{ 64 } -\dfrac { { (y-1) }^{ 2 } }{ 36 } =1$

So, option C is correct.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

If $PN$ is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, then the midpoint of $PN$ is

  1. circle

  2. parabola

  3. ellipse

  4. hyperbola
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $xy={ c }^{ 2 }$ be the rectangular hyperbola and let $P\left( { x
} _{ 1 },{ y } _{ 1 } \right) $ be apoint on it. Let $Q(h.k)$ be the
midpoint of $PN$. Then the coordinates of $Q$ are $\left( { { x } _{ 1
},{ y } _{ 1 } }/{ 2 } \right) $
$\therefore \quad { x } _{ 1 }={ h
}\quad \cfrac { { y } _{ 1 } }{ 2 } =k\Rightarrow { x } _{ 1 }={ h }\quad
,\quad { y } _{ 1 }=2k$
But $\left( { x } _{ 1 },{ y } _{ 1 } \right) $ lies on $xy={ c }^{ 2 }$
$\therefore \quad h(2k)={ c }^{ 2 }\Rightarrow hk=\cfrac { { c }^{ 2 } }{ 2 } $
Therefore, the locus of $(h,k)$ is $xy=\cfrac { { c }^{ 2 } }{ 2 } $, which is a hyperbola.
Hence, option 'D' is correct.

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The asymptotes of the hyperbola $xy - 3x + 4y + 2 = 0$ are

  1. $x= - 4$
  2. $x= 4$
  3. $y= - 3$
  4. $y= 3$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation
Given : Hyperbola,
$xy-3x+4y+2=0$---------------1
For Asymptotes,
Let the Asymptote's Equation be $y=mx+c$
And then finding $\phi _{n}(m)$ by replacing $y\rightarrow m$ and $x\rightarrow 1$
As $n=2$,
$\phi _{2}(m)=m$
putting $\phi _{2}(m)=0$, we get $m=0$
By taking $m=0$, we will get only one asymptote parallel to X-axis, so let's find them with putting the co-efficients of higher terms to zero.
For Asymptote parallel to X-axis, we put co-efficient of highest degree of x to zero that is here 1, so co-efficient of x$=0$
$\Rightarrow (y-3)=0$------------2(from Equation 1)
For Asymptote parallel to Y-axis, we put co-efficient of highest degree of y to zero which is 1 here, co-efficient of y$=0    (from Equation 1)
$\Rightarrow x+4=0$------------3
The Equation 2 & 3 are asymptotes to Equation 1.
$x+4=0$ & $y-3=0$




Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

Find the locus of the point of intersection of the lines $\sqrt 3 x-y-4\sqrt 3\lambda=0$ and $\sqrt 3 \lambda x +\lambda y-4\sqrt{3}=0$ for different values of $\lambda$.

  1. $3x^2-y^2=48$
  2. $y^2-3x^2=24$
  3. $4x^2-3y^2=16$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $(h,k)$ be the point of intersection of the given lines. Then,

$\sqrt 3 h-k-4\sqrt 3 \lambda=0$ and $\sqrt3 \lambda h +\lambda k-4\sqrt 3=0$
$\sqrt 3 h-k=4\sqrt 3\lambda$ and $\lambda(\sqrt 3h+k)=4\sqrt 3$
$(\sqrt 3 h-k)\lambda (\sqrt 3h +k)=(4\sqrt 3\lambda)(4\sqrt 3)$
$3h^2-k^2=48$
Hence, the locus of $(h,k)$ is $3x^2-y^2=48$.

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

If the equation of a hyperbola is $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{16}} = 1$, then 

  1. traverse axis is along x-axis of length $6$
  2. traverse axis is along y-axis of length $8$
  3. conjugate axis is along y-axis of length $6$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the hyperbola x^2/9 - y^2/16 = 1, a^2 = 9 and b^2 = 16. The transverse axis is along the x-axis with length 2a = 2 * 3 = 6.

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

For hyperbola  $\dfrac{x^2}{16}-\dfrac{y^2}{25}=1$ distances between two directrices are

  1. $\dfrac{16}{\sqrt{41}}$
  2. $\dfrac{25}{\sqrt{41}}$
  3. $\pm\dfrac{32}{\sqrt{141}}$
  4. $\dfrac{32}{\sqrt{41}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given equation of hyperbola $\dfrac {x^2}{16}-\dfrac {y^2}{25}=1$
Here $a^2=16, b^2=25$
Distance between directrix is  $\dfrac{2a^2}{\sqrt{a^2+b^2}} = \dfrac{2\times16}{\sqrt{16+25}} = \dfrac{32}{\sqrt{41}}$